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In some lecture notes, I saw the following procedure:

$$ \int_0^{\infty} x^{a-1} \int_x^{\infty} \exp(-t) (t-x)^{b-1} dt \, dx $$

is equal to:

$$ \int_0^{\infty} \int_0^{t} x^{a-1} \exp(-t) (t-x)^{b-1} dx \, dt $$

I did not understand the change of the limits of integrals. Thanks.

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This is an application of the theorems given in this Wikipedia article: en.wikipedia.org/wiki/Order_of_integration_(calculus). –  Stijn Nov 18 '12 at 18:18
    
Nevertheless, I am pretty sure that these theorems have nice proofs. :) Thanks. –  Mark Nov 18 '12 at 18:25
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1 Answer

up vote 4 down vote accepted

The integral $$\int_0^{\infty} \int_{x}^{\infty} f(x,y) dy dx$$ is evaluated as shown in the figure below.enter image description here

The integral $\displaystyle \int_{x}^{\infty} f(x,y) dy$ is first evaluated for a $dx$ red strip by letting $y$ go from $x$ to $+\infty$ and then the red strip is moved from $x=0$ to $x=\infty$. By doing this we have evaluated the integral $\displaystyle \int \int_{\Omega} f(x,y) dA$ where $\Omega$ is the blue region.

The second integral $$\int_0^{\infty} \int_{0}^{y} f(x,y) dx dy$$ is evaluated as shown in the figure below.enter image description here

The integral $\displaystyle \int_{0}^{y} f(x,y) dx$ is first evaluated for a $dy$ red strip by letting $x$ go from $0$ to $y$ and then the red strip is moved from $y=0$ to $y=\infty$. By doing this we have again evaluated the integral $\displaystyle \int \int_{\Omega} f(x,y) dA$ where $\Omega$ is the blue region.

Both the integrals must be the same since both evaluate the double integral over the blue area, $\Omega$.

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Excellent answer, thanks! –  Mark Nov 18 '12 at 19:00
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